(ii) To find the stationary point, set \(\frac{dy}{dx} = 0\):

\(\frac{2}{\sqrt{x}} - 1 = 0\)

\(\frac{2}{\sqrt{x}} = 1\)

\(\sqrt{x} = 2\)

\(x = 4\)

Substitute \(x = 4\) into the curve equation to find \(y\):

\(y = 6\)

Thus, the coordinates of the stationary point are \((4, 6)\).

(iii) Differentiate \(\frac{dy}{dx} = \frac{2}{\sqrt{x}} - 1\) to find \(\frac{d^2y}{dx^2}\):

\(\frac{d^2y}{dx^2} = -\frac{3}{x^2}\)

Since \(\frac{d^2y}{dx^2} < 0\), the stationary point is a maximum.

(iv) The gradient of the tangent at \(P\) is \(\frac{dy}{dx} = -\frac{1}{3}\).

The gradient of the normal is \(m = 3\) (since \(m_1 m_2 = -1\)).

The angle \(\theta\) is given by \(\tan \theta = 3\), so \(k = 3\).